Question

Two parallel plates

Equipotential Lines represent for us on a graph the same thing Elevation Lines do for us on a map: They represent a line where the Electric or Gravitational potential is a constant. Maps and voltage graphs have even spacing between values (1V or 500ft or 100m etc)so they can be read easier. These potentials were at 3V, 6V, 9V, 12V, and 15V. In the figure are 5 lines of electric potential drawn from positive charge at left and a negative charge at right.

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a)Draw tiny line segments perpendicular to your equipotential lines. See if you can connect a line segment perpendicular from one equipotential line to another line segment to another equipotential line keeping the ’curve’ needed to attach the two. Then see if you can connect that line to the next.

b) Connect this line to the conductor by making sure it comes perpendicularly out or into the conductor

c) This creates an electric field lined

d) Draw and label at least six electric field lines. Have them be roughly evenly spaced represent different portions of the map.

e) Indicate the direction of the electric field with an arrow on each of your electric field lines.

f) Measure the distance along the electric field “curve” using a string or other method to find the distance the electric field must travel between equipotential lines

g) Now the Electric Field is E=−∆V/∆s where ∆V is your separation of voltage between the two equipotential lines and ∆s is the distance the Electric Field curve traveled.

h) Find the Electric Field for at least 10 points on each map (a good representation of all different points on the map). To ensure as much precision as possible, only consider paths between adjacent equipotential lines. This means ∆V should be the same value for each calculation(your step value of 3V or 2V, etc)

Answer 1

Question a to e is in this picture:

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h)

−∆V = 3V as the final voltage is smaller than the initial voltage

at point 1, E=−∆V/∆s = -(9-12)/0.016 = 187.5 V/m

2> E=−∆V/∆s = 3/0.02 = 150V/m

3> E=−∆V/∆s = 3/0.021 = 142.85V/m

4> E=−∆V/∆s = 3/0.025= 120V/m

5> E=−∆V/∆s = 3/0.025 = 120V/m

6> E=−∆V/∆s = 3/0.015 = 200V/m

7> E=−∆V/∆s = 3/0.016 = 187.5V/m

8> E=−∆V/∆s = 3/0.015 = 197.3V/m

9> E=−∆V/∆s = 3/0.016 = 187.5 V/m

10> E=−∆V/∆s = 3/0.015= 187.5 V/m